使用并行数组编写简单的字符串哈希映射.重新哈希后获取某些键的 null 值

Question

我第一次put在方法中遇到冲突时，即当hasKey返回-1rehashing方法启动并且触发冲突的值变为双倍数组到可能的空插槽。但是System.out.println(m.get("1000"));给了我一些键的空，这意味着它们丢失了。我不明白它们是如何丢失的，因为没有什么可以覆盖它们keyArray.

import java.util.*;
public class StringMapParallel implements Iterable<String>{
private int nButckets = 2000;
private String[] keyArray = new String[nButckets];
private String[] valueArray = new String[nButckets];
private int numberOfEntries = 0;
public static void main(String[] args) {

StringMapParallel m = new StringMapParallel();
;
for (int i = 0; i < 8000; i++) {
m.put(String.valueOf(i), String.valueOf(i));
}
System.out.println(m.get("1000"));

}

private void rehashing() {
if (numberOfEntries > (int) (0.3 * nButckets)) {
int newBucketsNumber = nButckets * 2;
String[] newKeyArray = new String[newBucketsNumber];
String[] newValueArray = new String[newBucketsNumber];
for (int i = 0; i < keyArray.length; i++) {
if (keyArray[i] != null) {
int index = keyArray[i].hashCode() % newBucketsNumber;
newKeyArray[index] = keyArray[i];
newValueArray[index] = valueArray[keyArray[i].hashCode() % nButckets];
}
}           
/*
for (String key: this) {
int index = key.hashCode() % newBucketsNumber;
newKeyArray[index] = key;
if (key == null) System.out.println(key); 
newValueArray[index] = valueArray[key.hashCode() % nButckets];              
}
*/
keyArray = newKeyArray;
valueArray = newValueArray;
nButckets = newBucketsNumber;
}
}
public void put(String key, String value) {
int index = key.hashCode() % nButckets;
int hasKey = hasKey(index, key);
if (hasKey == 1) {
valueArray[index] = value;
} else if (hasKey == 0){
keyArray[index] = key;
valueArray[index] = value;
numberOfEntries++;
} else {
rehashing();
index = key.hashCode() % nButckets;
keyArray[index] = key;
valueArray[index] = value;
numberOfEntries++;
}
}
public String get(String key) {
int index = key.hashCode() % nButckets;
if (hasKey(index, key) == 1) return valueArray[index];
return null;
}
private int hasKey(int index, String key) {
if (keyArray[index] == null) {
return 0;
} else if (keyArray[index].equals(key)) {
return 1;
} else {
return -1;
}
}
public Iterator<String> iterator(){
Iterator<String> iter = new Iterator<String>() {
private int currentIndex = 0;
private int nEntries = 0;
@Override
public boolean hasNext() {
return nEntries < numberOfEntries && numberOfEntries != 0;
}
@Override
public String next() {
for (int i = currentIndex; i < keyArray.length; i++) {
if (keyArray[i] != null) {
currentIndex = i + 1;
nEntries++;
return keyArray[i];
}
}
return null;
}
};
return iter;
}
}

Answer 1

算法中键1000和6357的int index = key.hashCode() % nButckets;值相同，1423。因此，您的算法会用keyArray[1423]=6357覆盖keyArray[1423]=1000

当你打印m.get(String.valueOf(1000))时，按照get()方法检查index和key，因此它会返回null。阅读代码中的注释以获取进一步说明。

public String get(String key) {
//System.out.println(key+" "+ key.hashCode());
int index = key.hashCode() % nButckets;
System.out.println(index+"bfb"+hasKey(index, key));
//hasKey(index, key) would return -1, because key[1423] is 6357, and not 1000 as you expected.
if (hasKey(index, key) == 1) return valueArray[index]; 
return null;
}
private int hasKey(int index, String key) {
//System.out.println(keyArray[index]);
if (keyArray[index] == null) {
return 0;
} else if (keyArray[index].equals(key)) {
return 1;
} else {
return -1;
}
}